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Question Number 123141 by ajfour last updated on 23/Nov/20

Commented by ajfour last updated on 23/Nov/20

$$Findsidelengthofthecongruent$$$$squaresifBC=1andACsuch$$$$thattwosuchcongruentsquares$$$$bepossible.\mathrm{\angle }B=90°.$$

Answered by mr W last updated on 23/Nov/20

$$\mathrm{\angle }C=\theta$$$$AB=\tan \theta$$$$\frac{s}{\frac{1}{2}-\frac{s}{2}}=\frac{\frac{1}{2}\tan \theta -s}{\frac{s}{2}}$$$$\frac{2s}{1-s}=\frac{\tan \theta -2s}{s}$$$$\Rightarrow s=\frac{\tan \theta }{2+\tan \theta }$$$$similarly$$$$\Rightarrow s=\frac{\frac{1}{\tan \theta }}{2+\frac{1}{\tan \theta }}=\frac{1}{2\tan \theta +1}$$$$\frac{1}{2\tan \theta +1}=\frac{\tan \theta }{2+\tan \theta }$$$$\Rightarrow \tan \theta =1$$$$\Rightarrow AB=BC$$

Commented by ajfour last updated on 23/Nov/20

$$ThanksSir,sillyquestion.$$

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